D=1.2+3.4+4.5+...+19.20
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\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=1-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{1}-\frac{1}{20}\)
\(=\frac{19}{20}\)
\(D=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
=1-1/20
=19/20
a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=179/380
b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)
\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)
c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)
=1-1/21
=20/21
d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)
\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)
a) Ta có công thức tính tổng các số tự nhiên liên tiếp sau:
\(\Rightarrow1275=\frac{\left(1+n\right)n}{2}\Rightarrow\left(1+n\right)n=1275.2=2550=50.51\)
Mà n là số tự nhiên => n và n+1 là 2 số tự nhiên liên tiếp => n=50.
b) Đề chưa đầy đủ.
c) Ta có:
\(A=1.2+2.3+3.4+.....+19.20\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+19.20.\left(21-18\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+19.20.21-18.19.20\)
\(=\left(1.2.3+2.3.4+3.4.5+......+19.20.21\right)-\left(1.2.3+2.3.4+......+18.19.20\right)=19.20.21\)
\(\Rightarrow A=19.20.7=2660=133.2.10\Rightarrow\frac{A}{133.2}=\frac{2.133.10}{2.133}=10\)
cảm ơn bạn, mà đề chỉ là nếu có thôi chứ câu b đủ rồi á bạn
D=1.2+2.3+3.4+..+19.20
=>3D=1.2.3+2.3.3+3.4.3+...+19.20.3
=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)
=>1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
=19.20.21=7980
=>D=2660
vậy D=2660
\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\cdot\cdot\cdot+\dfrac{1}{18\cdot19}+\dfrac{1}{19\cdot20}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
#\(Urushi\)☕
D = 1.2+3.4+5.6+...+19.20
3D = 1.2(3-0) + 2.3(4-1) ...... 19.20 (21-18)
3D = 1.2.3-0.2.3.4-1 ......... 19.20.21-18
3D = ( 1.2.3+2.3.4+.....+19.20.21) - ( 0.1.2+ 1.2.3+.... + 18.19.20)
3D = 19.20.21
3D = 7980
D = 2660
D=1.2+3.4+4.5+...+19.20
3D=1.2.(3-0)+2.3.(4-1)+4.5(6-3)+...+19.20.(21-18)
3D=1.2.3-0.1.2+2.3.4-1.2.3+4.5.6-3.4.5+...+19.20.21-18.19.20
3D=(1.2.3+2.3.4+3.4.5+...+19.20.21)-(0.1.2+1.2.3+2.3.4+...+18.19.20)
3D=19.20.21
3D=7890
D=2660